### 3.146 $$\int \tanh ^2(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx$$

Optimal. Leaf size=63 $-\frac{b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac{(a+b)^2 \tanh (c+d x)}{d}+x (a+b)^2-\frac{b^2 \tanh ^5(c+d x)}{5 d}$

[Out]

(a + b)^2*x - ((a + b)^2*Tanh[c + d*x])/d - (b*(2*a + b)*Tanh[c + d*x]^3)/(3*d) - (b^2*Tanh[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0755494, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3670, 461, 206} $-\frac{b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac{(a+b)^2 \tanh (c+d x)}{d}+x (a+b)^2-\frac{b^2 \tanh ^5(c+d x)}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Tanh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

(a + b)^2*x - ((a + b)^2*Tanh[c + d*x])/d - (b*(2*a + b)*Tanh[c + d*x]^3)/(3*d) - (b^2*Tanh[c + d*x]^5)/(5*d)

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \tanh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-(a+b)^2-b (2 a+b) x^2-b^2 x^4+\frac{a^2+2 a b+b^2}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{(a+b)^2 \tanh (c+d x)}{d}-\frac{b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac{b^2 \tanh ^5(c+d x)}{5 d}+\frac{(a+b)^2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=(a+b)^2 x-\frac{(a+b)^2 \tanh (c+d x)}{d}-\frac{b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac{b^2 \tanh ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 0.0478517, size = 137, normalized size = 2.17 $\frac{a^2 \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{a^2 \tanh (c+d x)}{d}-\frac{2 a b \tanh ^3(c+d x)}{3 d}+\frac{2 a b \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{2 a b \tanh (c+d x)}{d}-\frac{b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^2 \tanh ^3(c+d x)}{3 d}+\frac{b^2 \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{b^2 \tanh (c+d x)}{d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Tanh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

(a^2*ArcTanh[Tanh[c + d*x]])/d + (2*a*b*ArcTanh[Tanh[c + d*x]])/d + (b^2*ArcTanh[Tanh[c + d*x]])/d - (a^2*Tanh
[c + d*x])/d - (2*a*b*Tanh[c + d*x])/d - (b^2*Tanh[c + d*x])/d - (2*a*b*Tanh[c + d*x]^3)/(3*d) - (b^2*Tanh[c +
d*x]^3)/(3*d) - (b^2*Tanh[c + d*x]^5)/(5*d)

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Maple [B]  time = 0.005, size = 189, normalized size = 3. \begin{align*} -{\frac{{a}^{2}\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{2\,d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) ab}{d}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ){b}^{2}}{2\,d}}-{\frac{{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-2\,{\frac{ab\tanh \left ( dx+c \right ) }{d}}-{\frac{2\,ab \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ){a}^{2}}{2\,d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) ab}{d}}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ){b}^{2}}{2\,d}}-{\frac{{a}^{2}\tanh \left ( dx+c \right ) }{d}}-{\frac{{b}^{2}\tanh \left ( dx+c \right ) }{d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x)

[Out]

-1/2*a^2/d*ln(tanh(d*x+c)-1)-1/d*ln(tanh(d*x+c)-1)*a*b-1/2/d*ln(tanh(d*x+c)-1)*b^2-1/3*b^2*tanh(d*x+c)^3/d-1/5
*b^2*tanh(d*x+c)^5/d-2*a*b*tanh(d*x+c)/d-2/3*a*b*tanh(d*x+c)^3/d+1/2/d*ln(tanh(d*x+c)+1)*a^2+1/d*ln(tanh(d*x+c
)+1)*a*b+1/2/d*ln(tanh(d*x+c)+1)*b^2-a^2*tanh(d*x+c)/d-b^2*tanh(d*x+c)/d

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Maxima [B]  time = 1.16304, size = 312, normalized size = 4.95 \begin{align*} \frac{1}{15} \, b^{2}{\left (15 \, x + \frac{15 \, c}{d} - \frac{2 \,{\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac{2}{3} \, a b{\left (3 \, x + \frac{3 \, c}{d} - \frac{4 \,{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + a^{2}{\left (x + \frac{c}{d} - \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/15*b^2*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) + 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c) + 45*e^(-8*d*x -
8*c) + 23)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d
*x - 10*c) + 1))) + 2/3*a*b*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2
*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) + a^2*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1)))

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Fricas [B]  time = 2.09198, size = 1245, normalized size = 19.76 \begin{align*} \frac{{\left (15 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \,{\left (15 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} -{\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (15 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 5 \,{\left (2 \,{\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} + 16 \, a b + 5 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (2 \,{\left (15 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (15 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \,{\left (15 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \,{\left ({\left (15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \,{\left (9 \, a^{2} + 16 \, a b + 5 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a^{2} + 8 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/15*((15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)^5 + 5*(15*(a^2 + 2*a*b + b^2)*d*x
+ 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 - (15*a^2 + 40*a*b + 23*b^2)*sinh(d*x + c)^5 + 5*(15
*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c)^3 - 5*(2*(15*a^2 + 40*a*b + 23*b^2)*cosh(d*
x + c)^2 + 9*a^2 + 16*a*b + 5*b^2)*sinh(d*x + c)^3 + 5*(2*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b
^2)*cosh(d*x + c)^3 + 3*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c))*sinh(d*x + c)^2
+ 10*(15*(a^2 + 2*a*b + b^2)*d*x + 15*a^2 + 40*a*b + 23*b^2)*cosh(d*x + c) - 5*((15*a^2 + 40*a*b + 23*b^2)*co
sh(d*x + c)^4 + 3*(9*a^2 + 16*a*b + 5*b^2)*cosh(d*x + c)^2 + 6*a^2 + 8*a*b + 10*b^2)*sinh(d*x + c))/(d*cosh(d*
x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c
))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

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Sympy [A]  time = 0.919852, size = 117, normalized size = 1.86 \begin{align*} \begin{cases} a^{2} x - \frac{a^{2} \tanh{\left (c + d x \right )}}{d} + 2 a b x - \frac{2 a b \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 a b \tanh{\left (c + d x \right )}}{d} + b^{2} x - \frac{b^{2} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac{b^{2} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac{b^{2} \tanh{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{2} \tanh ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)**2*(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Piecewise((a**2*x - a**2*tanh(c + d*x)/d + 2*a*b*x - 2*a*b*tanh(c + d*x)**3/(3*d) - 2*a*b*tanh(c + d*x)/d + b*
*2*x - b**2*tanh(c + d*x)**5/(5*d) - b**2*tanh(c + d*x)**3/(3*d) - b**2*tanh(c + d*x)/d, Ne(d, 0)), (x*(a + b*
tanh(c)**2)**2*tanh(c)**2, True))

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Giac [B]  time = 1.2481, size = 294, normalized size = 4.67 \begin{align*} \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (d x + c\right )}}{d} + \frac{2 \,{\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 45 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 180 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 220 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 140 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 140 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 70 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 40 \, a b + 23 \, b^{2}\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

(a^2 + 2*a*b + b^2)*(d*x + c)/d + 2/15*(15*a^2*e^(8*d*x + 8*c) + 60*a*b*e^(8*d*x + 8*c) + 45*b^2*e^(8*d*x + 8*
c) + 60*a^2*e^(6*d*x + 6*c) + 180*a*b*e^(6*d*x + 6*c) + 90*b^2*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) + 220*
a*b*e^(4*d*x + 4*c) + 140*b^2*e^(4*d*x + 4*c) + 60*a^2*e^(2*d*x + 2*c) + 140*a*b*e^(2*d*x + 2*c) + 70*b^2*e^(2
*d*x + 2*c) + 15*a^2 + 40*a*b + 23*b^2)/(d*(e^(2*d*x + 2*c) + 1)^5)